All categories

3 years ago

Will a cold snowball give off cold air?

Chemistry

1 answer:

rewona [7]3 years ago

6 0

No, Heat moves from concentrations of high to low.

You might be interested in

Solid substances are most likely to sublime if they have...?

tankabanditka [31]

Technically the are both right and wrong. Solid substances are most likely to sublime is they have... high vapor pressure and weak intermolecular forces.

5 0

3 years ago

Read 2 more answers

URGENT PLZ HELP AHHHHHHH

dexar [7]

A, B, and D. You already answered the first one, so I wouldn't think this would be too hard

8 0

3 years ago

The steps of scientific method in order are?

ArbitrLikvidat [17]

The answer is choice B

8 0

3 years ago

Read 2 more answers

Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro

horrorfan [7]

Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

Volume = Area (of the face) * Length

We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)

5 0

2 years ago

How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2

jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol$

For the given chemical reaction:

$BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)$

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = $\frac{1}{1}\times 0.00286mol=0.00286mol$ of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

$0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g$

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0

3 years ago